Thread: Quickie! Ambiguous Cases

After determing if a triangle is either case 1: a < 90 or case 2: a > 90.

In order to find how many solutions (if there is any) there is

EX:

A= 30 a= 13 c=26

Its case 1. Since its less than 90. After this do I do 26SIN(30)=____

To find how many solutions?
I know the guidelines for the solutions...

Originally Posted by topaz192

After determing if a triangle is either case 1: a < 90 or case 2: a > 90.

In order to find how many solutions (if there is any) there is

EX:

A= 30 a= 13 c=26

Its case 1. Since its less than 90. After this do I do 26SIN(30)=____

To find how many solutions?
I know the guidelines for the solutions...

There are 2 solutions... meaning 2 lengths for b, 2 angles for b, and 2 angles for c. I'm not really sure I really understand all of what you are saying, though.

Lol I dont either after reading. I guess I just dont understand how you know how to find how many solutions there are? How did you find it?

there is only one solution, and it's a right triangle.

Ok I get it now. Its one solution because its both 13 , right?

I kept thinking I was doing it wrong trying to get two solutions! lol

How do you find it if its an angle - angle - side??

EX:
A= 58
C= 94
b= 17

my teacher didnt go over ANYTHING like that!

Here,




So, there is only one solution, one triangle.

Originally Posted by topaz192

EX:
A= 58
C= 94
b= 17

my teacher didnt go over ANYTHING like that!

When you have two angles of a triangle you can determine the third. In your example, which I quoted above, B = 180 - A - C = 180 - 58 - 94 = 28. Then you have 17/sin(28) = c/sin(94) = a/sin(58).

So itd just be no solutions. since itll be less than the bsinA